# How do you find the center and vertices of the ellipse x^2/9+y^2/5=1?

##### 2 Answers
Jul 20, 2018

Please see the explanation below

#### Explanation:

Compare this equation to the standard equation of the ellipse

${\left(x - h\right)}^{2} / {a}^{2} + {\left(y - k\right)}^{2} / {b}^{2} = 1$

Our equation is

${x}^{2} / 9 + {y}^{2} / 5 = 1$

$a = \sqrt{9} = \pm 3$

$b = \pm \sqrt{5}$

The center is $C = \left(h , k\right) = \left(0 , 0\right)$

The vertices are :

$A = \left(h , a\right) = \left(3 , 0\right)$

$A ' = \left(h , - a\right) = \left(- 3 , 0\right)$

$B = \left(k , b\right) = \left(0 , \sqrt{5}\right)$

$B ' = \left(k , - b\right) = \left(0 , - \sqrt{5}\right)$

graph{(x^2/9+y^2/5-1)=0 [-7.9, 7.904, -3.95, 3.95]}

See the answers below

#### Explanation:

The given equation of ellipse:

${x}^{2} / 9 + {y}^{2} / 5 = 1$

${x}^{2} / {3}^{2} + {y}^{3} / \left\{{\left(\sqrt{5}\right)}^{2}\right\} = 1$

The above equation of ellipse is in form of ${x}^{2} / {a}^{2} + {y}^{2} / {b}^{2} = 1$ which has

Center: $\left(x = 0 , y = 0\right) \setminus \equiv \left(0 , 0\right)$

Vertices: (x=\pm a, y=0)\ \ &\ \ \ (x=0, y=\pmb)

(\pm3, 0) \ \ \ &\ \ \ (0, \pm\sqrt5)